Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I hope this helps. The explanation goes as: -We know angle CBA=35 -AOB=diameter and hence angle ACB=90 -Thus in triangle ACB, angle CAB=55 -Now in triangle AOC, OA=OC=Radius and thus angle ACO=55 and angle AOC=70. -Similarly we can find angle DOB=70 -Thus angle COD=180-70-70=40 And length of arc calculation is given in the image=2π

Can anyone please help me in solving this problem.......

There are 2-3 methods, lets see one of them:--- 1]You are given that OR is diameter=18 (radius =9); therefore arc OR is a semicircle the semicircle is composed of 3 arc--arc OP + arc PQ+ arc QR {arc PQ = 180- arc OP + arc QR}

2]the lines PQ & OR are parallel , therefore than alternate angles are congruent ( angle PRO = angle=QPR ) but these are inscribed angles and we need the central angles to compute the length of arc given by the formula : {central angle (teta) /360 * 2*pi*radius}

3] Central angle of arc QR or arc OP =2 * inscribed angle (2*35= 70) thus arc's QR + OP have central angles 70 +70= 140

4] thus, central angle for arc PQ =180-140=40 ;

length of arc PQ= 40/360 *2* pi* 9 = 2 pi

You may also employ back solving if you are comfortable with that for faster solution _________________

Rules for posting on the verbal forum When you post a question Pls. Provide its source & TAG your questions Avoid posting from unreliable sources such as 1000 series.

In the circle above, PQ is parallel to diameter OR [#permalink]

Show Tags

10 May 2012, 20:40

Here I'm giving an elaborate solution. Hope it helps. Let A be the center of the circle. We join the mid point of PQ, let's say S with A. Clearly angle ASQ=90 degree. Now angle ROQ= angle PRO= 35 degree As at center any arc subtends double the angle what it subtends at any peripherial point, we can conclude angle RAQ= 70 degree. Hence angle PAQ= 2* angle SAQ= 2*(90-70)= 40 degree 180 degree= pi radian 40 degree= 2 pi/9 rad Hence minor arc PQ= 9*2 pi/9 (As radius of the circle=9) Arc PQ= 2 pi Ans-(A)

In the circle above, PQ is parallel to diameter OR, and OR has [#permalink]

Show Tags

30 Jul 2012, 16:03

1

This post received KUDOS

I encountered this problem on one of the tests. It was around my 12th or 13 question. I do admit that this question stumped me and I took some amount of time in solving this question.

I started off with the geometric approach but then quickly got tangled in it. I then shifted to approximating the area and using the process of elimination. I am explaining my approach even though it is unconventional and based only on the situation that I was in.

assumption pi=3.14

The radius is given as (OR/2)=9. Hence the circumference of the circle is 18pi.

Clearly looking at the figure, the length of arc PQ is much smaller than 1/4 the circumference.

18*pi=18*3.14~42+ (read it as something greater than 42)

Hence 1/4 of the circumference is ~ 10+ (read as something greater than 10). The length of Arc PQ is much smaller than 1/4 of the circumference as well.

Options C and D can be eliminated since C=~10.5 and D=~13.5. These can be clearly eliminated. Similarly Option E : 3pi ~ very close to 10 and hence I eliminated it.

After this it was purely on the figure given than I eliminated the other choices. The arc PQ looks to be more like closer to half of the 1/4 i.e. ~5+. I really have no justification as to how I deduced it. Just an observation so that I could guess the answer.

Option B [(9pi)/4] comes to be roughly 7- (read as less than 7)

Option A (2pi) comes to ~6.5- (read as less than 6.5) which is closer to mu initial estimate of being 5+.

Hence I chose A.

I am very well aware that on some other day it is very much possible to choose B and to get the question wrong. Also I am not advocating this method in any way

Its just that this was an educated guess which I guessed had 50% chance of being right.

The above explanations and diagramatic representations have helped me realize my mistake and should such a problem appear again I am dead sure of the mathematical way to solve it. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

Re: In the circle above, PQ is parallel to diameter OR [#permalink]

Show Tags

12 Feb 2014, 03:56

1

This post received KUDOS

Bingo! central angle theorem is the rock star! Thanks for the valuable Math book Bunuel!

One other thing I missed in this problem is it asks for the minor arc length PQ. NOT length of PQ.

Follow up question. Is it possible to find length of the chord PQ with the information given in the question? If so what is the length of PQ? _________________

Re: In the circle above, PQ is parallel to diameter OR [#permalink]

Show Tags

23 Jun 2014, 00:41

Bunuel wrote:

In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ?

A. \(2\pi\)

B. \(\frac{9\pi}{4}\)

C. \(\frac{7\pi}{2}\)

D. \(\frac{9\pi}{2}\)

E. \(3\pi\)

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Let C be the center of the circle.

According to the central angle theorem above <PCO=2<PRO=70.

As PQ is parallel to OR, then <QPR=<PRO=35. Again, according to the central angle theorem above <QCR=2<QPR=70.

<PCQ=180-(<PCO+<QCR)=180-70-70=40.

Minor arc \(PQ=\frac{40}{360}*circumference=\frac{2\pi{r}}{9}=2\pi\)

Answer: A.

For more on circle check the circles chapter of Math Book (link in my signature).

Bunnel, don't you think that this question asks for a little too much imagination to be on GMAT? Just curious, would you rate this as a 750 range question on the test?

Re: In the circle above, PQ is parallel to diameter OR [#permalink]

Show Tags

23 Jun 2014, 01:48

Expert's post

Kconfused wrote:

Bunuel wrote:

In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ?

A. \(2\pi\)

B. \(\frac{9\pi}{4}\)

C. \(\frac{7\pi}{2}\)

D. \(\frac{9\pi}{2}\)

E. \(3\pi\)

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Let C be the center of the circle.

According to the central angle theorem above <PCO=2<PRO=70.

As PQ is parallel to OR, then <QPR=<PRO=35. Again, according to the central angle theorem above <QCR=2<QPR=70.

<PCQ=180-(<PCO+<QCR)=180-70-70=40.

Minor arc \(PQ=\frac{40}{360}*circumference=\frac{2\pi{r}}{9}=2\pi\)

Answer: A.

For more on circle check the circles chapter of Math Book (link in my signature).

Bunnel, don't you think that this question asks for a little too much imagination to be on GMAT? Just curious, would you rate this as a 750 range question on the test?

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...